What is the general solution for 3sin(x)+1=2csc(x) ?

The general solution for 3sin(x)+1=2csc(x) is x=(3pi)/2+2pin,x=0.72972…+2pin,x=pi-0.72972…+2pin

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3sin(x)+1=2csc(x)

Solution

3 sin (x) + 1 = 2 csc (x)

Solution

x = \frac{3 π}{2} + 2 πn, x = 0.72972 \dots + 2 πn, x = π - 0.72972 \dots + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n, x = 41.81031 \dots^{\circ} + 36 0^{\circ} n, x = 138.18968 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 sin (x) + 1 = 2 csc (x)

Subtract

2 csc (x)

from both sides

3 sin (x) + 1 - 2 csc (x) = 0

Rewrite using trig identities

1 - 2 csc (x) + 3 sin (x)

Use the basic trigonometric identity:

sin (x) = \frac{1}{csc ( x )}

= 1 - 2 csc (x) + 3 \cdot \frac{1}{csc ( x )}

3 \cdot \frac{1}{csc ( x )} = \frac{3}{csc ( x )}

3 \cdot \frac{1}{csc ( x )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 3}{csc ( x )}

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3}{csc ( x )}

= 1 - 2 csc (x) + \frac{3}{csc ( x )}

1 + \frac{3}{csc ( x )} - 2 csc (x) = 0

Solve by substitution

1 + \frac{3}{csc ( x )} - 2 csc (x) = 0

Let:

csc (x) = u

1 + \frac{3}{u} - 2 u = 0

1 + \frac{3}{u} - 2 u = 0 : u = - 1, u = \frac{3}{2}

1 + \frac{3}{u} - 2 u = 0

Multiply both sides by

u

1 + \frac{3}{u} - 2 u = 0

Multiply both sides by

u

1 \cdot u + \frac{3}{u} u - 2 uu = 0 \cdot u

Simplify

1 \cdot u + \frac{3}{u} u - 2 uu = 0 \cdot u

Simplify

1 \cdot u : u

1 \cdot u

Multiply:

1 \cdot u = u

= u

Simplify

\frac{3}{u} u : 3

\frac{3}{u} u

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 u}{u}

Cancel the common factor:

u

= 3

Simplify

- 2 uu : - 2 u^{2}

- 2 uu

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= - 2 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= - 2 u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

u + 3 - 2 u^{2} = 0

u + 3 - 2 u^{2} = 0

u + 3 - 2 u^{2} = 0

Solve

u + 3 - 2 u^{2} = 0 : u = - 1, u = \frac{3}{2}

u + 3 - 2 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 u^{2} + u + 3 = 0

Solve with the quadratic formula

- 2 u^{2} + u + 3 = 0

Quadratic Equation Formula:

For

a = - 2, b = 1, c = 3

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

1^{2} - 4 (- 2) \cdot 3 = 5

1^{2} - 4 (- 2) \cdot 3

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 3

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 ( - 2 )}, u_{2} = \frac{- 1 - 5}{2 ( - 2 )}

u = \frac{- 1 + 5}{2 ( - 2 )} : - 1

\frac{- 1 + 5}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 2}

Add/Subtract the numbers:

- 1 + 5 = 4

= \frac{4}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- 1 - 5}{2 ( - 2 )} : \frac{3}{2}

\frac{- 1 - 5}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 5}{- 2 \cdot 2}

Subtract the numbers:

- 1 - 5 = - 6

= \frac{- 6}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 6}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{6}{4}

Cancel the common factor:

2

= \frac{3}{2}

The solutions to the quadratic equation are:

u = - 1, u = \frac{3}{2}

u = - 1, u = \frac{3}{2}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

1 + \frac{3}{u} - 2 u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 1, u = \frac{3}{2}

Substitute back

u = csc (x)

csc (x) = - 1, csc (x) = \frac{3}{2}

csc (x) = - 1, csc (x) = \frac{3}{2}

csc (x) = - 1 : x = \frac{3 π}{2} + 2 πn

csc (x) = - 1

General solutions for

csc (x) = - 1

csc (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

csc (x) = \frac{3}{2} : x = arccsc (\frac{3}{2}) + 2 πn, x = π - arccsc (\frac{3}{2}) + 2 πn

csc (x) = \frac{3}{2}

Apply trig inverse properties

csc (x) = \frac{3}{2}

General solutions for

csc (x) = \frac{3}{2}

csc (x) = a \Rightarrow x = arccsc (a) + 2 πn, x = π - arccsc (a) + 2 πn

x = arccsc (\frac{3}{2}) + 2 πn, x = π - arccsc (\frac{3}{2}) + 2 πn

x = arccsc (\frac{3}{2}) + 2 πn, x = π - arccsc (\frac{3}{2}) + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn, x = arccsc (\frac{3}{2}) + 2 πn, x = π - arccsc (\frac{3}{2}) + 2 πn

Show solutions in decimal form

x = \frac{3 π}{2} + 2 πn, x = 0.72972 \dots + 2 πn, x = π - 0.72972 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3sin(x)+1=2csc(x) ?
The general solution for 3sin(x)+1=2csc(x) is x=(3pi)/2+2pin,x=0.72972…+2pin,x=pi-0.72972…+2pin